 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-2/19/c/iii/solution

For every fixed real

a

, the base is positive once

x > ∣ a ∣

. Taking its logarithm gives

lo g [(\frac{x + a}{x - a})^{x}] = x [lo g (1 + \frac{a}{x}) - lo g (1 - \frac{a}{x})] .

2 a / x + O (x^{- 3})

, so this logarithm tends to

2 a

. Continuity of the exponential gives

x \to \infty lim (\frac{x + a}{x - a})^{x} = e^{2 a} .

For

a = 0

the base is identically one, consistently giving the same answer.

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